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Question-74371




Question Number 74371 by liki last updated on 23/Nov/19
Commented by liki last updated on 23/Nov/19
....plz help me 16a ,,emergerce
$$….{plz}\:{help}\:{me}\:\mathrm{16}{a}\:,,{emergerce} \\ $$
Answered by MJS last updated on 23/Nov/19
ky=kx+x+7  y=x+(x/k)+(7/k)  y=x(1+(1/k))+(7/k)  1+(1/k)=2 ⇒ k=1  ⇒ y=2x+7
$${ky}={kx}+{x}+\mathrm{7} \\ $$$${y}={x}+\frac{{x}}{{k}}+\frac{\mathrm{7}}{{k}} \\ $$$${y}={x}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)+\frac{\mathrm{7}}{{k}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{{k}}=\mathrm{2}\:\Rightarrow\:{k}=\mathrm{1} \\ $$$$\Rightarrow\:{y}=\mathrm{2}{x}+\mathrm{7} \\ $$
Commented by liki last updated on 23/Nov/19
...Thanks sir
$$…{Thanks}\:{sir} \\ $$
Answered by peter frank last updated on 24/Nov/19
ky=kx+x+7  ky^′ =k+1  y^′ =2  2k=k+1  k=1  y−intercept x=0  ky=7  y=7
$${ky}={kx}+{x}+\mathrm{7} \\ $$$${ky}^{'} ={k}+\mathrm{1} \\ $$$${y}^{'} =\mathrm{2} \\ $$$$\mathrm{2}{k}={k}+\mathrm{1} \\ $$$${k}=\mathrm{1} \\ $$$${y}−{intercept}\:{x}=\mathrm{0} \\ $$$${ky}=\mathrm{7} \\ $$$${y}=\mathrm{7} \\ $$

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